How do you prove that the function $f\left( x \right) = {\left( {x + 2{x^3}} \right)^4}$is continuous at $a = - 1$?

Hint: To solve this question, we need to determine two things in order to show that the function is continuous at the given point. First, the limit of the given function as it approaches to the given point if it exists. Second, the given function’s value at the given points. By using these values, we can determine whether the function is continuous or not at the given point.

Complete step by step solution:
We are given the function $f\left( x \right) = {\left( {x + 2{x^3}} \right)^4}$ and a point $a = - 1$.
First we will find the limit of the function $f\left( x \right) = {\left( {x + 2{x^3}} \right)^4}$ as it approaches the given point $a = - 1$.
This can be written as:
\[
  \mathop {\lim }\limits_{x \to a} f\left( x \right) \\
   = \mathop {\lim }\limits_{x \to - 1} {\left( {x + 2{x^3}} \right)^4} \\
   = {\left( { - 1 + 2{{\left( { - 1} \right)}^3}} \right)^4} \\
   = {\left( { - 3} \right)^4} \\
   = 81 \\
 \]
Now, we will find the value of the function at the given point which is $f\left( a \right)$.
$
  f\left( x \right) = {\left( {x + 2{x^3}} \right)^4} \\
   \Rightarrow f\left( a \right) = {\left( {a + 2{a^3}} \right)^4} \\
   \Rightarrow f\left( a \right) = {\left( {a + 2{a^3}} \right)^4} = {\left( { - 1 + 2{{\left( { - 1} \right)}^3}} \right)^4} = {\left( { - 3} \right)^4} = 81 \\
 $
Here, we can observe that the value of the limit of the function $f\left( x \right) = {\left( {x + 2{x^3}} \right)^4}$ as it approaches to the given point and the value of the function at the given point $a = - 1$.
That is \[\mathop {\lim }\limits_{x \to - 1} f\left( x \right) = f\left( { - 1} \right)\].
Therefore, we can say that the function $f\left( x \right) = {\left( {x + 2{x^3}} \right)^4}$ is continuous at $a = - 1$.

Note: In this question, we have proved that the function is continuous at the given point mathematically. However, we can also understand it graphically with a simple logic. A graph for a function that is smooth without any holes, jumps, or asymptotes is called the continuous function.